\(A=1.3+3.5+5.7+...+45.47+47.49\)

\(A=\left(1.49\right)+\left(2.3\right)+\left(2.5\right)+\left(2.7\right)+.....+\left(2.47\right)\)

\(A=49+2.\left(3+5+7+....+47\right)\)

Bây giờ ta phải tìm SSH của :

\(3+7+...+47\)

Vậy SSH của tổng đó là :

(47-3):2+1=23 (SSH)

=> \(A=49+2.\left(\frac{\left(47+3\right).23}{2}\right)\)

\(A=49+2.575\)

\(A=49+1150\)

\(A=1199\)

Dạng này lầm đầu gặp

\(6A=1.3.6+3.5.6+5.6.7+......+47.49.6=3+1.3.5+3.5.\left(7-1\right)+5.7.\left(9-3\right)+.....+47.49.\left(51-45\right)=3+1.3.5-1.3.5+3.5.7-......+47.49.51-45.47.49=47.49.51+3=3+141423=141426\Rightarrow A=23571\)

ngồi cả chục phút bấm máy tính ( Dùng hệ đếm cơ số N cho dễ nhin )

VÀ ĐÂY LÀ KẾT QUẢ ( ĐÃ KIỄM TRA KO SAI MỘT BIỂU THỨC , KO TIN AE CÓ THỂ DÙNG MÁY TÍNH KIỂM TRA )

1/1-1/3+1/3-1/5+1/5-1/7+...... +1/47-1/49

 3/1.3+3/3.5+3/5.7+......+3/47.49

=1/1-1/3+1/3-1/5+1/5-1/7+........+1/47-1/49

=1/1-1/49

=49/49-1/49

=48/49

= 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 +......+ 1/47 - 1/49

= 1 - 1/49

= 48/49 nha!

*** Ai k mk mk k lại !!***

\(2H=\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{49.51}\)

\(2H=\dfrac{3-1}{1.3}+\dfrac{5-3}{3.5}+...+\dfrac{51-49}{49.51}\)

\(2H=\dfrac{3}{1.3}-\dfrac{1}{1.3}+\dfrac{5}{3.5}-\dfrac{3}{3.5}+...+\dfrac{51}{49.51}-\dfrac{49}{49.51}\)

\(2H=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{49}-\dfrac{1}{51}\)

\(2H=1-\dfrac{1}{51}\)

\(2H=\dfrac{50}{51}\)

\(H=\dfrac{25}{51}\)

\(F=\dfrac{49}{2.9}+\dfrac{49}{9.16}+............+\dfrac{49}{65.72}\)

\(\Leftrightarrow F=\dfrac{7^2}{2.9}+\dfrac{7^2}{9.16}+............+\dfrac{7^2}{65.72}\)

\(\Leftrightarrow F=7\left(\dfrac{7}{2.9}+\dfrac{7}{9.16}+.............+\dfrac{7}{65.72}\right)\)

\(\Leftrightarrow F=7\left(\dfrac{1}{2}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+...........+\dfrac{1}{65}-\dfrac{1}{75}\right)\)

\(\Leftrightarrow F=7\left(\dfrac{1}{2}-\dfrac{1}{72}\right)\)

\(\Leftrightarrow F=7.\dfrac{35}{72}=\dfrac{245}{72}\)

\(G=\dfrac{3}{1.3}+\dfrac{3}{3.5}+...........+\dfrac{3}{47.49}\)

\(\Leftrightarrow G=\dfrac{3.2}{1.3.2}+\dfrac{3.2}{3.5.2}+........+\dfrac{3.2}{47.49}\)

\(\Leftrightarrow G=\dfrac{3}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+..........+\dfrac{2}{47.49}\right)\)

\(\Leftrightarrow G=\dfrac{3}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+........+\dfrac{1}{47}-\dfrac{1}{49}\right)\)

\(\Leftrightarrow G=\dfrac{3}{2}\left(1-\dfrac{1}{49}\right)\)

\(\Leftrightarrow G=\dfrac{3}{2}.\dfrac{48}{49}=\dfrac{72}{49}\)

\(C=\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{45.47}\)

\(C=\frac{5}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{45}-\frac{1}{47}\right)\)

\(C=\frac{5}{2}.\left(1-\frac{1}{47}\right)\)

\(C=\frac{5}{2}.\frac{46}{47}\)

\(C=\frac{115}{47}\)

= 205/101 nha

\(C=\frac{5}{1\cdot3}+\frac{5}{3\cdot5}+\frac{5}{5\cdot7}+...+\frac{5}{45\cdot47}\)

\(C=\frac{5}{2}\cdot\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{45\cdot47}\right)\)

\(C=\frac{5}{2}\cdot\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{45}-\frac{1}{47}\right)\)

\(C=\frac{5}{2}\cdot\left(1-\frac{1}{47}\right)\)

\(C=\frac{5}{2}\cdot\frac{46}{47}\)

\(C=\frac{115}{47}\)

theo công thức, ta tính đc: 

A = 1- 1/3 + 1/3 - 1/5 + 1/5 -1/7 +..... + 1/49 - 1/51

=> A bằng 1- 1/51 ( các cặp phân số đối nhau thì lược bỏ như - 1/3 và + 1/3 )

1-1/3=2/3 chứ ko phải 1-1/3=1/3 đâu nha bạn

theo bài ra ta có:

A=1-1/3+1/3-1/5+1/5-1/7+......+1/47-1/49+1/49-1/51

A=1-1/51

\(\Leftrightarrow\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{47}-\frac{1}{49}\right)=\frac{1}{x}\)

\(\Leftrightarrow\frac{1}{2}\left(1-\frac{1}{49}\right)=\frac{1}{x}\Rightarrow x=\frac{49}{24}\)

\(\frac{1}{2}.\left(1-\frac{1}{3}\right)+\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}\right)+\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{7}\right)+...+\frac{1}{2}.\left(\frac{1}{47}-\frac{1}{49}\right)=\frac{1}{x}\)

\(\frac{1}{2}.\left(1-\frac{1}{49}\right)=\frac{1}{x}\)

\(\frac{24}{49}=\frac{1}{x}\)\(\Rightarrow x=\frac{49}{24}\)

có ai giúp đc mik k